Need Help with Vector Problem? Confused About Velocity and Angle?

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The discussion focuses on solving a vector problem involving a boat crossing a river with a current. The river flows east at 1.50 m/s, while the boat maintains a velocity of 10.0 m/s at an angle of 35 degrees northwest relative to the water. The correct velocity of the boat relative to the shore is calculated to be 8.81 m/s at an angle of 139 degrees, and the boat moves 582.8 meters downstream by the time it reaches the north shore. Participants highlight confusion regarding the interpretation of angles and the distinction between velocity and speed.

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Help with Vector problem??

A river flows due east at 1.50m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant velocity of 10.0m/s 35 degrees northwest relative to the water. a) What is the velocity of the boat relative to the shore? b) If the river is 500m wide, how far downstream has the boat moved by the time it reaches the north shore?

Hi, I the answers to this problem is 8.81 @ 139 degrees and 582.8m
I tried to do this, and was successful with finding the velocity of the boat, but for some reason my angle is off. I keep getting 40.63. Can anyone explain to me what I did wrong? And for the second part, I don't have a clue how to start it. Thanks!

I took a picture of my work:
http://tinypic.com/r/2d14k1t/6
 
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Well for part 2, the river is flowing horizontally, so it does not affect the component of the boat's velocity perpendicular to the shore. You can act as though the river was still to get the amount of time it takes to get from one shore to the other. Since you want the horizontal distance, you take the horizontal component of the boat's velocity times the time.
 


I see you interpreted "35 deg NW " as 35 deg north of west whereas I would interpret it as 35 deg west of north. But in any case I didn't get the "right" answers either.

The wording is also confusing regarding "constant velocity of 10 m/s relative to the water". The term "velocity" should I think be "speed". I would then interpret that to mean
Let vx = east component of ground velocity of boat
vy = north component of ground velocity of boat
Then (vx - 1.5)^2 + vy^2 = 10^2 = 100.

Then also, vx/vy = tan 35 deg.

Would others please join in? :rolleyes:
 
Last edited:

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